My question is given as:
$\gamma \ : \ [0,\pi] \rightarrow \mathbb{C}$ for the function $\gamma(t)=2e^{i6t}$, compute the path integral $\int_\gamma z^5 dz$
My attempt so far is $$\begin{align}\int_\gamma z^5 dz & = \int_0^\pi \gamma(t)^5 \gamma'(t) dt \\ &= \int_0^\pi (2e^{i6t})^5 (12ie^{i6t})dt \\ & = 384i \int^\pi_0 e^{36it} dt \\ & = 384i \bigg[\frac{-1}{36}ie^{36it}\bigg]^\pi_0 \\ & = 384i \bigg(\frac{-1}{36}i--\frac{1}{36}i \bigg) \\ & = 0 \end{align}$$
I don't believe this is correct because I got 0, could anyone clarify if what I have done is correct?
Help is greatly appreciated.
According to Cauchy's integral theorem, this is entirely expected. It says that if $f \colon \mathbb{C} \to \mathbb{C}$ is holomorphic, and $\gamma$ is a simple closed curve, then $$\oint_\gamma f\,dz = 0$$.